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jbritto
Group Title
What is the limit of (1+cosx)/sinx when x approaches 0
 one year ago
 one year ago
jbritto Group Title
What is the limit of (1+cosx)/sinx when x approaches 0
 one year ago
 one year ago

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zugzwang Group TitleBest ResponseYou've already chosen the best response.0
l'Hôpital's rule, are you allowed to use that?
 one year ago

zugzwang Group TitleBest ResponseYou've already chosen the best response.0
Wait, I panicked... no l'Hôpital's rule, I'm sorry
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
How can you use lhospital here?
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
You have 2/0 not 0/0
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
\[\lim_{x \rightarrow 0}\frac{1+\cos(x)}{\sin(x)}\] Recall the following: \[\lim_{x \rightarrow 0}\frac{\sin(x)}{x}=1 \text{ , } \lim_{x \rightarrow 0}\frac{\cos(x)1}{x}=0\] Let's see if we can use that here: \[\lim_{x \rightarrow 0}\frac{\cos(x)+1}{\sin(x)} \cdot \frac{\cos(x)1}{\cos(x)1}\] \[\lim_{x \rightarrow 0} \frac{\cos^2(x)1}{\sin(x)(\cos(x)1)}\] \[\lim_{x \rightarrow 0}=\frac{\sin^2(x)}{\sin(x)(\cos(x)1)}\] \[\lim_{x \rightarrow 0}\frac{\sin(x)}{\cos(x)1}\] Now try it :)
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
Using L'hospital you will get something easier to look at :)
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
We didn't actually need any of those things I told you to recall above :)
 one year ago

jbritto Group TitleBest ResponseYou've already chosen the best response.0
Thanks. It still becomes 1/0 which is undefined ??
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
its not 1/0 but its 0/0 @jbritto sin0 = 0 cos0 =1
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
anyways,,there is an alternative also.. 1+ cosx = 1 + 2 cos^2 (x/2) 1 = 2cos^2 (x/2) and sinx = 2 sin(x/2) cos(x/2) so whole expression is 2 cos^2(x/2) / 2 sin(x/2) cos(x/2) = cos(x/2)/ sin(x/2) = cot (x/2) not plug in x=0
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
oh,,so you were saying it becomes 1/0 !! ohh,,am sorry.. in that case either the limit is undefined or you maybe made a typing error//
 one year ago

jbritto Group TitleBest ResponseYou've already chosen the best response.0
yes. applying LHosp rule, it will become cosx/sinx which is 1/0 undefined
 one year ago
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